Is $\mathbb{Z}[G]$ integral over $\mathbb{Z}$? Here $G$ is a finite group(not neccessarily abelian),then there is a statement in some representation book that $\mathbb{Z}[G]$ is integral over $\mathbb{Z}$.That is, every element in $\mathbb{Z}[G]$ satisfies a monic polynomial equation with coefficients in $\mathbb{Z}$.
How to get this result?
I worked with the case $G=S_3$ and found it is indeed this case, and I know it also holds for the abelian case trivially, yet I have no idea how to get the general result.
Will someone be kind enough to give me some hints on this?Thank you very much!
 A: Marc's argument doesn't work as written; the obvious map $\mathbb{Z}[S_n] \to \mathcal{M}_n(\mathbb{Z})$ isn't injective. Indeed the latter is a free $\mathbb{Z}$-module on $n^2$ generators while the former is a free $\mathbb{Z}$-module on $n!$ generators... 
Fortunately, there's an easy way out. $\mathbb{Z}[G]$ acts faithfully on itself by left multiplication ("Cayley's theorem for rings"), and this directly defines an injection $\mathbb{Z}[G] \to \mathcal{M}_{|G|}(\mathbb{Z})$. 
A: Corrected in response to comment by @Qiaochu Yuan.
Embed $G$ into the symmetric group $S_n$ for $n=\#G$  using Cayley's theorem (don't take a shortcut if $G$ is already a permutation group), and map the ring $\mathbf{Z}[S_n]$ homomorphically to the matrix ring $M_n(\mathbf{Z})$ using permutation matrices. Although the second map is not injective, the composed map $\mathbf{Z}[G]\to M_n(\mathbf{Z})$ is, as can be seen by looking at the first column.
Now apply the Cayley-Hamilton theorem.
A: The subset $A$ of $\Bbb Z[G]$ of integers elements over $\Bbb Z$ is a subring.
You want to show that $A = \Bbb Z[G]$.
Thus is it enough to prove that the (canonical) generators of $\Bbb Z[G]$ are integers over $\Bbb Z$.
This is trivial since they are all root of unity : for $g\in G$, $g^{|G|} = 1$.
Remark — The determinant trick presented in the other answers is often used to show that the subset of integer elements is a subring. 
